Bài 4: 

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

hay O,B,A,C cùng thuộc 1 đường tròn

Bài 5:

\(\sqrt{x+2021}-y^3=\sqrt{y+2021}-x^3\\ \Leftrightarrow\left(\sqrt{x+2021}-\sqrt{y+2021}\right)+\left(x^3-y^3\right)=0\\ \Leftrightarrow\dfrac{x-y}{\sqrt{x+2021}+\sqrt{y+2021}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\\ \Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-y=0\\\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2=0\left(1\right)\end{matrix}\right.\)

Dễ thấy \(\left(1\right)>0\) với mọi x,y

Do đó \(x-y=0\) hay \(x=y\)

\(\Leftrightarrow M=x^2+2x^2-2x^2+2x+2022=x^2+2x+1+2021\\ \Leftrightarrow M=\left(x+1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=-1\)

Bài 1.

a. $=a^2+2.a.12+12^2=a^2+24a+144$

b. $=(3a)^2+2.3a.\frac{1}{3}+(\frac{1}{3})^2=9a^2+2a+\frac{1}{9}$

c. $=(5a^2)^2+2.5a^2.6+6^2=25a^4+60a^2+36$

d. $=\frac{1}{4}+2.\frac{1}{2}.4b+(4b)^2$

$=\frac{1}{4}+4b+16b^2$

e.

$=(a^m)^2+2.a^m.b^n+(b^n)^2$

$=a^{2m}+2a^mb^n+b^{2n}$

Bài 2.

$(x-0,3)^2=x^2-0,6x+0,09$

$(6x-3y)^2=36x^2-36xy+9y^2$

$(5-2xy)^2=25-20xy+4x^2y^2$
$(x^4-1)^2=x^8-2x^4+1$

$(x^5-y^3)^2=x^{10}-2x^5y^3+y^6$

Bài 3.

a. $(x+10)^2+(x-10)^2=x^2+20x+100+(x^2-20x+100)$

$=2x^2+200$

b. $(x-12)^2+(x+12)^2=x^2-24x+144+(x^2+24x+144)$

$=2x^2+288$

c. $(x+7)^2-(x-7)^2=[(x+7)+(x-7)][(x+7)-(x-7)]$

$=2x.14=28x$

Bài 5 ở đâu vậy ạ?

\(a,\dfrac{6.12-6.7}{60}=\dfrac{6\left(12-7\right)}{60}=\dfrac{6.5}{60}=\dfrac{30}{60}=\dfrac{1}{2}\\ b,\dfrac{35.18-35}{\left(-34\right).7}=\dfrac{35\left(18-1\right)}{\left(-34\right).7}=\dfrac{35.17}{\left(-34\right).7}=\dfrac{595}{-238}=\dfrac{-5}{2}\)

\(c,\dfrac{42-11.42}{21.\left(-15\right)}=\dfrac{42\left(1-11\right)}{21.\left(-15\right)}=\dfrac{2.21.\left(-10\right)}{21.\left(-15\right)}=\dfrac{-20}{-15}=\dfrac{4}{3}\)

tách hình ra chứ để z chả ai rảnh làm hết đâu :vv

l: Ta có: \(\sqrt{x^2-10x+25}=5-x\)

\(\Leftrightarrow\left|x-5\right|=5-x\)

\(\Leftrightarrow x-5\le0\)

hay \(x\le5\)

\(a,ĐK:x\ge5\\ PT\Leftrightarrow2\sqrt{x-5}+3\cdot\dfrac{1}{3}\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=4\\ \Leftrightarrow4\sqrt{x-5}=4\\ \Leftrightarrow x-5=1\Leftrightarrow x=6\left(tm\right)\\ b,ĐK:x\ge1\\ PT\Leftrightarrow\dfrac{2}{3}\cdot3\sqrt{x-1}-\dfrac{1}{4}\cdot4\sqrt{x-1}+27\cdot\dfrac{1}{9}\sqrt{x-1}=4\\ \Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\\ c,ĐK:2\le x\le8\\ PT\Leftrightarrow x-2=x^2-16x+64\\ \Leftrightarrow x^2-17x+66=0\\ \Leftrightarrow\left(x-11\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=11\left(ktm\right)\\x=6\left(tm\right)\end{matrix}\right.\Leftrightarrow x=6\)

\(d,ĐK:x\ge8\\ PT\Leftrightarrow x-2=x^2-16x+6x\left(làm.như.câu.c\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=11\left(tm\right)\\x=6\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=11\)

\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\)

<=>\(\dfrac{4\left(7x-2\right)}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3\left(x-2\right)}{12}\)

<=>\(4\left(7x-2\right)-24x< 60-3\left(x-2\right)\)

<=>\(28x-8-24x< 60-3x+6\)

<=>\(28x+3x-24x< 60+8+6\)

<=>\(7x< 74\)

<=>x<\(\dfrac{74}{7}\)

Vậy...

\(W=\dfrac{1}{2}\left(\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\right)\\ W=\dfrac{1}{2}\left(2+\sqrt{2}+2-\sqrt{2}\right)=\dfrac{1}{2}\cdot4=2\\ Y=\dfrac{1}{2}\left(\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(4-\sqrt{3}\right)^2}\right)\\ Y=\dfrac{1}{2}\left(4+\sqrt{3}+4-\sqrt{3}\right)=\dfrac{1}{2}\cdot8=4\)

có thể tự viết ạ, dàn ý đầy đủ lắm bạn cần thêm quan hệ từ, từ nối, liên kết các câu cho chặt chẽ thế là xong:3